When graphing and describing the characteristics of a parabola, it is important to know several key pieces of information. The parabola intercepts describe where the parabola intersects the x-axis and the y-axis while the vertex of a parabola is the highest (or lowest) point of the parabola. Knowing the domain and range of a parabola is also helpful when graphing.

Finding the intercepts domain and rangeand another way to find the vertexa parabola.

So whenever we're finding X intercepts,what we want to do is let Y is equalto 0. So we let Y equal 0.What we end up with is 0 is equal to X squaredminus 5X plus 6. So what weend up with is a quadratic equation,which we have the tools to solve.

We can either factor, complete the square,quadratic formula, a number of differentways to solve this out. This particular examplefactors quite easily. We end up with X minus2. X minus 3. So we know that our intercepts on R2and 3. That's easy enough for ourX intercepts.

For our Y intercepts, what we want todo is let X equal 0. When we letX equals 0 our first term disappears,our second term disappears, andwe're just left with our constant termwhich in this case is just goingto be 6.

Finding the vertex.So two ways of finding the vertex.We can either complete the square, whichtends to be pretty involved, or thereis a little shortcut which is negative Bover 2A is equal to the X coordinate.of the vertex.

So all we have to do is go to our equation.Remember the coefficient on X squaredis A. Coefficient on X is B andthe constant term is C. So negativeB over 2A in this case is justgoing to be negative, negative 5, 5 over2 times A which is just 2 times1. So we find out that our X coordinateis 5 over 2. Okay.

In order to find the other part of our vertex,we just found the X coordinate.We still have to find the Y. All we haveto do is plug in the X coordinateinto this equation.

This particular example isn't the nicest,because we're dealing with a fractionbut it still doesn't matter all we haveto do is plug in two and a half.We end up with two and a half squared minus5 times 2.5 plus 6 and we end upwith negative 1.25. So by plugging in that negative B over 2A into the equation we end up with ourY coordinate of our vertici.

To find the domain, domain is value ofX that we have to put in, there's norestrictions. We're not dividing by 0. There's no square roots.So X can actually be whatever it wants.So we can call it all reals,negative infinity topositive infinity, different ways of sayingthe same thing.

The last thing we're looking at is the rangeof the Y values. This is going to take a littlebit of piecing together.So we know we have a parabola.Coefficient on the X squared term is 1,which means our parabola is going tobe facing upwards.

So we know we have an upwardfacing parabola.And we know that our vertex isat a point 5/2s and 1/4.So the Y coordinate of our vertex is going to be the lowest pointupward facing parabola.Y coordinate at the bottom.So what we end up is our range being fromnegative 1/4 to infinity.

Our range actually hits that point.That point is actuallya point on the curve.So we then can include negative1/4 and we're going up.If this was a negative coefficient on theX squared term, I know that the parabolawould be going down.

So I would know that I'm going from negativeinfinity all the way up to thatY coordinate of the vertex.So X intercept and Y intercept are similarto finding any other X intercept,Y intercept, let the other coordinateequal 0 and solve.

Little trick for finding the vertex.Negative V 2 over A plug it in to findthe Y coordinate and domain and rangeand domain is going to be all realliesfor parabola, range just considerthe vertex and if the graphis going up or down.